package leetcode;

/**
 * 130. 被围绕的区域
 * 给你一个 m x n 的矩阵 board ，由若干字符 'X' 和 'O' ，找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。
 * <p>
 * <p>
 * 示例 1：
 * <p>
 * <p>
 * 输入：board = [["X","X","X","X"],["X","O","O","X"],["X","X","O","X"],["X","O","X","X"]]
 * 输出：[["X","X","X","X"],["X","X","X","X"],["X","X","X","X"],["X","O","X","X"]]
 * 解释：被围绕的区间不会存在于边界上，换句话说，任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上，或不与边界上的 'O' 相连的 'O' 最终都会被填充为 'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。
 * 示例 2：
 * <p>
 * 输入：board = [["X"]]
 * 输出：[["X"]]
 */
public class Solve {

    /**
     * 一看就是回溯方法，往四个方向找
     *
     * @param board
     */
    public void solve(char[][] board) {

        if (board == null || board.length == 0) {
            return;
        }

        int m = board.length;
        int n = board[0].length;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                boolean isEdge = i == 0 || i == m - 1 || j == 0 || j == n - 1;
                if (isEdge && board[i][j] == 'O') {
                    dfs(board, i, j);
                }
            }
        }

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                }
                if (board[i][j] == 'A') {
                    board[i][j] = 'O';
                }
            }
        }
    }

    // O -> A
    public void dfs(char[][] board, int i, int j) {
        if (i < 0 || i >= board.length || j < 0 || j >= board[0].length) {
            return;
        }
        if (board[i][j] == 'X' || board[i][j] == 'A') {
            return;
        }

        board[i][j] = 'A';
        dfs(board, i - 1, j);
        dfs(board, i + 1, j);
        dfs(board, i, j - 1);
        dfs(board, i, j + 1);
    }
}
